Question

In a metal deficient oxide sample, M_XY₂O₄(M and Y are metals), M is present in both +2 and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M²⁺ ions present in M is \(\frac{1}{3}\), the value of X is______.

• A. 0.25
• B. 0.33
• C. 0.67
• D. 0.75

Answer 1

The Correct Option is D

Charge Neutrality and Fraction of Ions Calculation 

Consider the compound \( \text{M}_x \text{Y}_2 \text{O}_4 \), where the M ions exist in two oxidation states: \( \text{M}^{2+} \) and \( \text{M}^{3+} \). The fraction of \( \text{M}^{2+} \) ions is given as \( \frac{1}{3} \). Thus:

\[ \text{M}^{2+} = \frac{x}{3}, \quad \text{M}^{3+} = \frac{2x}{3} \]

Step 1: Apply Charge Neutrality

For the compound to be neutral, the total positive and negative charges must balance. This gives us the equation: \[ \frac{2x}{3} \cdot (+3) + \frac{x}{3} \cdot (+2) + 2 \cdot (+3) + 4 \cdot (-2) = 0 \] Simplify the terms: \[ \frac{6x}{3} + \frac{2x}{3} + 6 - 8 = 0 \]

Step 2: Solve for \( x \)

Combine terms: \[ \frac{6x}{3} + \frac{2x}{3} = 2 \] Multiply through by 3 to eliminate the denominators: \[ 6x + 2x = 6 \] Simplify: \[ 8x = 6 \quad \Rightarrow \quad x = \frac{6}{8} = 0.75 \]

Final Answer:

The value of \( x \) is \( x = 0.75 \).